How do you write an equation of a circle with center (3,0) passes through the point (-2,4)?

1 Answer
Shwetank Mauria
Dec 26, 2016

x^2+y^2-6x-32=0

Explanation:

WE know that the equation of a circle with center (a,b) and radius r is (x-a)^2-(y-b)^2=r^2.

Here although center is given as (3,0), its radius has not been given explicitly. But as circle passes through (-2,4) and radius is the distance between center and any point on the circumference, the radius is equal to distance between (3,0) and (-2,4), which is

sqrt((3-(-2))^2-(0-4)^2)=sqrt((3+2)^2-(-4)^2)=sqrt(25+16)=sqrt41

Hence equation of circle is

(x-3)^2-(y-0)^2=(sqrt41)^2

or x^2-6x+9+y^2=41

or x^2+y^2-6x-32=0
graph{x^2+y^2-6x-32=0 [-16, 16, -8, 8]}

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